# Solving Trig Equations

• Nov 24th 2007, 02:57 PM
Macleef
Solving Trig Equations
Solve each equation for $0 \leq \theta \leq 360$

--
$sin^2x - 1 = cos^2x$
--

Work:
$cos^2x - cos^2x = 0$
$0 = 0$

--
90 and 270
--

Incorporate the use of the double angle formulas identities and other identities to solve the following on $0 \leq \theta \leq 360$

--
$sin2x = 2cos2x$
--

Work:
$2sinxcosx - 2cosx = 0$
$2cosx (sinx - 1) = 0$

What do I do now?

--
90 and 270
--

Btw, how would you simply $cos^2x$ to $cosx$ if you had something like $cos^2x = 1$? Would it be...
$cosx = \sqrt{1}$
$x = cos^-1 \sqrt{1}$?
• Nov 24th 2007, 03:10 PM
topsquark
Quote:

Originally Posted by Macleef
--
$sin^2x - 1 = cos^2x$
--

Work:
$cos^2x - cos^2x = 0$
$0 = 0$

Not quite. $1 - sin^2(x) = cos^2(x)$, so
$sin^2x - 1 = cos^2x$

$-cos^2(x) = cos^2(x)$

$2cos^2(x) = 0$

$cos(x) = 0$

Thus $x = 90^o , 270^o$

-Dan
• Nov 24th 2007, 03:23 PM
topsquark
Quote:

Originally Posted by Macleef
Incorporate the use of the double angle formulas identities and other identities to solve the following on $0 \leq \theta \leq 360$

--
$sin2x = 2cos2x$
--

Work:
$2sinxcosx - 2cosx = 0$
$2cosx (sinx - 1) = 0$

What do I do now?

This is one of the nastier ones:
$sin(2x) = 2cos(2x)$

$2sin(x)cos(x) = 2cos(2x)$

$sin(x) cos(x) = cos(2x)$

$sin(x) cos(x) = 2cos^2(x) - 1$

$cos(x) (\pm)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1$

$cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1$

$5cos^4(x) - 5cos^2(x) + 1 = 0$

This has solutions
$cos(x) = \pm 0.85065080835215 \implies x = 31.7175^o, 148.283^0$
and
$cos(x) = \pm 0.52573111211905 \implies x = 58.2825^o, 121.717^o$

Of these, only $x = 31.7175^o, 121.717^o$ solve the original equation.

-Dan
• Nov 24th 2007, 03:25 PM
topsquark
Quote:

Originally Posted by Macleef
Btw, how would you simply $cos^2x$ to $cosx$ if you had something like $cos^2x = 1$? Would it be...
$cosx = \sqrt{1}$
$x = cos^-1 \sqrt{1}$?

How do you solve any equation where you take the square root of both sides?

$cos^2(x) = 1$

$cos(x) = \pm 1$

Thus
$x = cos^{-1}(1) = 0^o$
and
$x = cos^{-1}(-1) = 180^o$

-Dan
• Nov 24th 2007, 03:59 PM
Macleef
Quote:

Originally Posted by topsquark

This is one of the nastier ones:
$sin(2x) = 2cos(2x)$

$2sin(x)cos(x) = 2cos(2x)$

$sin(x) cos(x) = cos(2x)$

$sin(x) cos(x) = 2cos^2(x) - 1$

$cos(x) (\pm)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1$

$cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1$

$5cos^4(x) - 5cos^2(x) + 1 = 0$

This has solutions
$cos(x) = \pm 0.85065080835215 \implies x = 31.7175^o, 148.283^0$
and
$cos(x) = \pm 0.52573111211905 \implies x = 58.2825^o, 121.717^o$

Of these, only $x = 31.7175^o, 121.717^o$ solve the original equation.

-Dan

What the heck...? I'm so confused
I'm pretty sure this equation isn't that complicated...
• Nov 24th 2007, 04:01 PM
Macleef
Quote:

Originally Posted by topsquark
Not quite. $1 - sin^2(x) = cos^2(x)$, so
$sin^2x - 1 = cos^2x$

$-cos^2(x) = cos^2(x)$

$2cos^2(x) = 0$

$cos(x) = 0$

Thus $x = 90^o , 270^o$

-Dan

How did you get $-cos^2x$? Isn't it $cos^2x$ for $sin^2x - 1$?
• Nov 25th 2007, 07:26 AM
topsquark
Quote:

Originally Posted by Macleef
What the heck...? I'm so confused
I'm pretty sure this equation isn't that complicated...

This equation is. Whatever your book (teacher, professor, whatever) intended for you to solve certainly wasn't given the listed solution set. There's a typo in your given equation, but I can't figure out what it was meant to be.

-Dan
• Nov 25th 2007, 07:27 AM
topsquark
Quote:

Originally Posted by Macleef
How did you get $-cos^2x$? Isn't it $cos^2x$ for $sin^2x - 1$?

$sin^2(x) + cos^2(x) = 1$

$sin^2(x) + cos^2(x) - 1 = 0$

$sin^2(x) - 1 = -cos^2(x)$

-Dan